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This puzzle was created by me as a single-player variation on the game of go. The objective here is simple: capture all 41 grey stones on the board below in the least number of moves.

The board.

Just like in the game of go, capturing a stone happens when a stone has no more liberties. Each stone has 4 liberties that must be taken (as shown in the image below).

Image courtesy of wikipedia.

You are given four stones, each a different colour: red, green, blue and violet. The order in which you place these 4 stones is determined by the "colour wheel" below. First red, then green, then blue, then violet, then red again, etc. Skipping a colour still counts as one move. Remember that you only have four stones; so when you go to place a new stone, you will be removing it from it's previous position.

The four colours.

Additional Rules:

  • You can place your stones anywhere you wish (including on top of other stones or even suicide).
  • You cannot lose your stones (in the case where you suicide).

You can post your answer as a list of x, y coordinates of where you placed your stones. If you skipped a colour, treat it as though you moved the stone to the same spot (i.e. write down the coordinate of where the stone is currently at).

Sadly I don't have a program right now that can simulate this puzzle within your browser. I would however be willing to offer a bounty if someone could provide me with one :)

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  • $\begingroup$ I think it would be much more interesting puzzle if you had 4 stones and you could move any stone any time. So you could optimize which stone you move and try to get the next gray where the most stone are in place already. $\endgroup$
    – user13856
    Commented Jul 6, 2015 at 19:46
  • $\begingroup$ @Gabor The only problem is that writing down all the moves might make things a little messy. If someone can provide me with a program that simulates this puzzle, I'll most likely add this in as a bonus. Thanks for the suggestion :) $\endgroup$
    – Allan
    Commented Jul 6, 2015 at 20:57

1 Answer 1

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My attempt gives 100 moves.

Start in one corner. Move along the diagonal, taking each grey stone with only 2 moves. Than move along an edge to the next grey stone (4 moves). Then move diagonal towards your original starting corner. Then repeat until you have removed half of the stones. Than re-position the stones above/beside your original starting position and repeat again.

I.E: (Starting at bottom left)
[(1,10)(0,9)(1,8)(2,9)] 4 moves - 1 grey stone   Total: 4 moves
[(2,7)(3,8)] 2 moves - 1 grey stone              Total: 6 moves
[(3,6)(4,7)] 2 moves - 1 grey stone              Total: 8 moves
....
[(9,0)(10,1)] 2 moves - 1 grey stone             Total: 20 moves

Now down one stone (4 moves) and diagonal to the other side (12 moves) etc.

In total it should take [(8*2)+(6*2)+(4*2)+(2*2)+(4*5)] for half and [(6*2)+(4*2)+(2*2)+(4*4)] for the other half, which gives us 100 moves in total.

A note from Allan: Here is a visual of the answer that I myself had come up with.

enter image description here

This is essentially the same as what Mark here is describing; it just does things in a different order.

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  • $\begingroup$ This is the answer that I got as well. I will hold off from giving you the check for the time being to see if someone else can give a better answer. $\endgroup$
    – Allan
    Commented Jul 6, 2015 at 19:16
  • $\begingroup$ Surely this can be optimized. In order to capture the first stone in the next diagonal, you've moved 4. You could have captured a stone in that same diagonal by moving only 3. There's probably other optimizations, too. To make the point even more clear, your very last capture only needs to move green, purple, and blue, but you move red anyway. (I understand that the rules still count "leaving red where it is" as one move, but I think that's ridiculous.) $\endgroup$
    – Ian MacDonald
    Commented Jul 6, 2015 at 20:06
  • $\begingroup$ @IanMacDonald I noticed about the 3 moves, but it might mess up your orientation for shifting to the next line. (I did not look into it too much) $\endgroup$
    – Mark N
    Commented Jul 6, 2015 at 20:09
  • $\begingroup$ Ah. Let's just focus on the last one (since optimizing one at a time is easier for me to think about). Consider the last four stones; if you start that next row with red and purple inverted (red on 2,7; purple on 0,7), then you'll end up with purple on 2,9. This means you can move red, green, blue to capture 1,9. $\endgroup$
    – Ian MacDonald
    Commented Jul 6, 2015 at 20:13

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